You yourself proved P2 true — fishfry

No I didn't.

Your argument is analogous to this:

If I am immortal then when will I die of old age? I won't. Therefore, I have proved that I am immortal.

Agreeing with what follows if we can recite the natural numbers at successively halved intervals of time doesn't prove that we can recite the natural numbers at successively halved intervals of time.

To make this extremely clear to you, as it seems I must, you must prove that we can recite the natural numbers at successively halved intervals of time.

It is the very premise that we can halve time ad infinitum (and perform some action within this time frame) that is being questioned. If you want to prove that supertasks are metaphysically possible then you must prove that this is metaphysically possible. You haven't done so.

But you just proved P2 yourself! You agreed that under the hypothesis of being able to recite a number at successively halved intervals of time, there is no number that is the first to not be recited. — fishfry

I agreed that if P2 is true then C1 is true, as I have agreed from the beginning.

This doesn't prove that P2 is true.

In favour of eternalism there's the relativity of simultaneity and the Rietdijk–Putnam argument.

I have agreed repeatedly that we can't "count all the natural numbers backwards" since an infinite sequence has no last element. — fishfry

So we're back to my post here:

a. I said "0", 30 seconds after that I said "1", 15 seconds after that I said "2", 7.5 seconds after that I said "3", and so on ad infinitum

b. I said "0", 30 seconds before that I said "1", 15 seconds before that I said "2", 7.5 seconds before that I said "3", and so on ad infinitum

Here is our premise:

P1. In both (a) and (b) there is a bijection between the series of time intervals and the series of natural numbers and the sum of the series of time intervals is 60.

However, the second supertask is metaphysically impossible. It cannot start because there is no largest natural number to start with. Therefore, P1 being true does not entail that the second supertask is metaphysically possible.

Therefore, P1 being true does not entail that the first supertask is metaphysically possible.

You accept that (b) is impossible but you claim that (a) is possible. You have to prove this. P1 doesn't prove it.

Given P2, what is the first natural number not recited? I seem to remember having asked you this several times already. — fishfry

There isn't one. I've answered this several times already. That's what it means for me to accept P1.

But you need to prove P2. You haven't done so.

People disagree with the premise because we are not confident we can use such intuitions when the — unintuitive — concept of infinity is involved. — Lionino

The important part from that post is this:

The fallacy in his reasoning is that it does not acknowledge that for all t_n >= t_1/2 the lamp is on iff the lamp was off and I pressed the button to turn it on and the lamp is off iff the lamp was on and I pressed the button to turn it off.

If the lamp is on at t₁ then it must have been either turned on at t₁ or turned and left on before t₁, neither of which are allowed given the supertask, hence the contradiction.

Me and fishfry have insisted that this is a case of missing limit. — Lionino

That's why it's impossible to complete.

You think that the end of the sequence at t=1 is a temporal/logical consequence of what happens before. — Lionino

Yes, I address that here.

Any given example does not prove that supertasks in general are necessarily impossible. — Lionino

I addressed this here and here.

True. And that implies time is discrete how? — Lionino

If time is continuous then supertasks are logically possible. Supertasks are logically impossible. Therefore, time is discrete.

I'm not trying to find a solution, just to understand what's going on. Not so much why it's wrong, but why anyone would think it was right. Where does the illusion come from? — Ludwig V

They're clearly being confused by maths. They think that because a geometric series of time intervals can have a finite sum and because this geometric series has the same cardinality as the natural numbers then it is possible to recite the natural numbers in finite time. Their conclusion is a non sequitur, and this is obvious when we consider the case of reciting the natural numbers (or any infinite sequence) in reverse.

There is a far more fundamental problem, and they're just ignoring it. I have no idea why. Perhaps because they can't look beyond the maths to what it would mean for us to actually carry out the tasks. This seems to be the mistake that Benacerraf made in his response to Thomson and which I addressed here.

It isn't a physically possible task. — noAxioms

It's not just physically impossible, it's logically impossible. No physics can allow me to begin reciting the natural numbers in reverse. I can't even say one number, let alone all of them. And this is true even if we're not reciting the natural numbers in reverse but the sequence {0, 1, 0, 1, ...} in reverse, i.e. where each term, individually, can be recited in less than a second.

That there is no first number to recite is the very reason that it is logically impossible to begin reciting them in reverse and it astonishes me that not only can't you accept this but you twist it around and claim that it not having a first number is the reason that it can begin without a first number.

Though I don't quite see how your B2 follows from your B1. — Ludwig V

It was redefined as 1 at t_1/2 and never changed again, so is still defined as 1 at t₁.

You mean that we don't know the state of X at the last step before t(1), even though X must have been in one state or the other? — Ludwig V

There is no last step before t₁, hence no coherent definition of X at t₁. But also at no point between t₀ and t₁ is there a step where X goes from being defined (as either "0" or "1") to being undefined, and the definition of X is always retained until redefined to something else. It's a simple contradiction.

If you're trying to find a "solution" you won't find one. We just have to accept that supertasks are illogical. It's that easy.

This puzzles me. Is this t(1) the same t as the t(1) in C3? It can't be. There must be a typo there somewhere. — Ludwig V

No, it was three separate situations. Sorry if that wasn’t clear.

One question, then - The state of X at any t(n), depends on its predecessor state at t(n-1), doesn't it? Isn't that a definition? Why is it inapplicable to t(1)? — Ludwig V

It is applicable to t₁, but given the supertask described in P3 there’s no coherent answer to the definition of X at t₁ (no final redefinition before t₁) proving P3 to be impossible.

I have wondered whether one could replace the Thompson lamp with a question, such as whether the final number was odd or even. That would work if you start with an odd divisor and don't express everything in decimals. Perhaps it would work for all examples if you ask whether the number of steps taken is odd or even when the minute is up. I think. — Ludwig V

P1. When the letter X is given a definition it retains this definition until it is redefined.

A1. At t₀ X = 0
A2. Therefore, at t₁ X = 0

B1. At t₀ X = 0 and then at t_1/2 X = 1
B2. Therefore, at t₁ X = 1

C1. At t₀ X = 0 and then at t_1/2 X = 1 and then at t_3/4 X = 0, and so on ad infinitum
C2. Therefore, at t₁ X = ?

In all cases the definition of X at t₁ must be a logical consequence of what occurs between t₀ and t₁.

Given that in C2 X cannot be defined as either "0" or "1" but must be defined as either "0" or "1" then C1 is necessarily false. The supertask described in C1 is impossible.

This addresses the very logic of a supertask without some dependency on a physical performance.

Consider the infinite sequence {0, 1, 0, 1, 0, 1, ...}.

Now consider reciting its terms in reverse.

To recite its terms in reverse I am only allowed to say "0" or "1" but I cannot start by saying "0" and I cannot start by saying "1". Therefore I cannot start.

No appeal to a geometric series of time intervals can save you from this.

The lack of a first step does not prevent the beginning of the task — noAxioms

It literally does.

I described exactly how to do that — noAxioms

No you didn't. You ignored it and just said "when the time comes I say the next number". That doesn't explain how the recitation can begin without a first number to say.

I am right now trying to recite the natural numbers in descending order but am silent because I cannot begin. It's been 60 seconds. Not only have I failed to recite them all, but I have failed to recite even one. Help me out here.

According to this, "many philosophers have argued that relativity implies eternalism. Philosopher of science Dean Rickles says that, "the consensus among philosophers seems to be that special and general relativity are incompatible with presentism." Christian Wüthrich argues that supporters of presentism can salvage absolute simultaneity only if they reject either empiricism or relativity."

Ok. Just talking about standard mathematical sequences. It's a common misunderstanding in this thread. The sequence 1/2, 3/4, 7/8, ... has a limit, namely 1, but no last element.

The sequence 1/2, 1/4, 1/8, ... also has a limit, namely 0, and no last element. But if you put the elements of the sequence on the number line, they appear to "come from" 0 via a process that could never have gotten started. This is my interpretation of Michael's example of counting backwards. — fishfry

This is what I mean by reciting backwards:

If I recite the natural numbers <= 10 backwards then I recite 10, then 9, then 8, etc.
If I recite the natural numbers <= 100 backwards then I recite 100, then 99, then 98, etc.

If I recite all the natural numbers backwards then I recite ... ?

It's self-evidently impossible. There's no first (largest) natural number for me to start with.

Given your reluctance to clarify the definition of the verb 'to start', I cannot respond appropriately to this statement. I gave a pair of options, or you can supply your own, so long as it isn't open to equivocation. — noAxioms

Just the ordinary meaning of "start", e.g. "begin".

You ask me, right now, to recite the natural numbers in descending order. How do I begin to perform this?

I think it's self-evident that I cannot begin because there is no first (largest) number for me to begin with.

I've repeatedly challenged you to name the first number not verbalized when we count forward 1, 2, 3, ... at successively halved intervals of time. — fishfry

I accept this:

P1. If we can recite forward 1, 2, 3, ... at successively halved intervals of time then we can recite all natural numbers in finite time

But I reject these:

P2. We can recite forward 1, 2, 3, ... at successively halved intervals of time
C1. We can recite all natural numbers in finite time

If you want to claim that C1 is true then you must prove that P2 is true. You haven't done so.

I think Thomson's lamp and similar examples prove that P2 is false. See here.

I cannot start reciting the natural numbers in descending order because there is no first natural number for me to start with.

That a geometric series has a finite sum is irrelevant to this very simple self-evident fact.

You (as well as Meta above) seem to insist on an additional premise of the necessity of a bound to something explicitly defined to be unbounded. — noAxioms

No, I'm saying that something with no start cannot start and something with no end cannot end.

Your argument is effectively "by definition it has no start therefore it can start without a start." You're trying to take the very thing that makes it impossible as proof that it's possible.

Good! Then it's logically possible for it to. An infinite number of things can complete without blowing up logic. — fdrake

But we're talking about supertasks, not geometric series. That a geometric series is possible isn't that a supertask is possible.

Given that there is no largest natural number it is logically impossible to even start reciting all the natural numbers in descending order.

I don't know why you think the existence of a geometric series proves otherwise.

How does it start? That's easy. When the appropriate time comes, the number to be recited at that time is recited. That wasn't so hard, was it? It works for both scenarios, counting up or down. — noAxioms

There is no first natural number to start with. It is logically impossible to have started reciting the natural numbers in descending order.

So when I hear Michael talking about the impossibility of a geometric series "completing" (so to speak) due to being unable to recite the terms in finite time... — fdrake

I don't think it impossible for a geometric series to complete. I think it impossible to have recited every natural number in descending order.

My issue is with supertasks, not with maths.

Repeating yourself three times, while ignoring my responses, does not help further the conversation. — Bob Ross

Your responses do not address my claim hence why I have to repeat it.

This is false; and does not follow from the former claim you made. — Bob Ross

It does follow. My belief that aliens exist makes the proposition "I believe that aliens exist" true. Therefore, your claim that "a belief cannot make a proposition true or false" is false.

Nah. That's an appeal to metaphysical or physical impossibility. Not logical impossibility! — fdrake

It is logically impossible to have recited every natural number in descending order because it is logically impossible to even start such a task.

So you’re claiming that it’s logically possible to have recited the natural numbers in descending order. That’s evidently absurd.

Which would be odd, seeing as such an object has a model in set theory — fdrake

Does it? Consider these two supertasks:

a. I said "0", 30 seconds after that I said "1", 15 seconds after that I said "2", 7.5 seconds after that I said "3", and so on ad infinitum

b. I said "0", 30 seconds before that I said "1", 15 seconds before that I said "2", 7.5 seconds before that I said "3", and so on ad infinitum

The first is reciting every natural number in ascending order and the second is reciting every natural number in descending order.

Does the second have a model in set theory? Is the second logically possible?

That there is a bijection between the series of time intervals and the series of natural numbers and that the sum of the series of time intervals is 60 says nothing about the possibility of (b) and so says nothing about the possibility of (a) either.

I think you're missing Bob Ross's point.

A belief that "aliens exist" is not the same as a belief about the proposition "I believe that aliens exist" — ChrisH

I haven't claimed otherwise.

I have only claimed this:

"I believe that aliens exist" is true iff I believe that aliens exist. Therefore his conclusion that "a belief cannot make a proposition true or false" is false.

Ok. I asked for a reference. Now I have no idea what I'm supposed to conclude from this. — fishfry

I said this:

Can you prove that it's metaphysically possible for me to halve the time between each subsequent recitation ad infinitum? It's not something that we can just assume unless proven otherwise. Even Benacerraf in his criticism of Thomson accepted this. — Michael

You responded with this:

Feel free to give a reference, else I can't respond. — fishfry

I gave you this reference:

I think it should be clear that, just as Thomson did not establish the impossibility of super-tasks by destroying the arguments of their defenders, I did not establish their possibility by destroying his.

So I ask again: can you prove that it's metaphysically possible for me to halve the time between each subsequent recitation ad infinitum?

Maybe I'm not being clear, so I'll try one more time.

Here are two proposed supertasks:

a. I said "0", 30 seconds after that I said "1", 15 seconds after that I said "2", 7.5 seconds after that I said "3", and so on ad infinitum

b. I said "0", 30 seconds before that I said "1", 15 seconds before that I said "2", 7.5 seconds before that I said "3", and so on ad infinitum

Here is our premise:

P1. In both (a) and (b) there is a bijection between the series of time intervals and the series of natural numbers and the sum of the series of time intervals is 60.

However, the second supertask is metaphysically impossible. It cannot start because there is no largest natural number to start with. Therefore, P1 being true does not entail that the second supertask is metaphysically possible.

Therefore, P1 being true does not entail that the first supertask is metaphysically possible.

If you want to argue that the first supertask can end despite there being no largest natural number to end with, and so is metaphysically possible, then you need something other than P1 to prove it.

you are thinking that "aliens exist" is true or false relative to a belief — Bob Ross

No I'm not.

I'm only saying that "I believe that aliens exist" is true iff I believe that aliens exist. Therefore your conclusion that "a belief cannot make a proposition true or false" is false.

Yes that is a proposition, and whether or not it is true or false is independent of any belief about it — Bob Ross

As per Tarski’s T-schema, “P” is true iff P. As such, “I believe that aliens exist” is true iff I believe that aliens exist.

Might show it's logically possible tho. — fdrake

You think that the super task I described might be logically possible? How would you start it?

By providing a standard mathematical object which is infinite, has no final element, tends to an end state, and has an infinite number of occurrences ("steps"), but occurs in finite time. — fdrake

I’ll repeat something I said above:

The fact that there is a bijection between the series of time intervals and the series of natural numbers and that the sum of the series of time intervals is 60 does not prove that the following supertask is metaphysically possible:

I said "0", 30 seconds before that I said "1", 15 seconds before that I said "2", 7.5 seconds before that I said "3", and so on ad infinitum.

A clock ticks 1 time per second.
You start with a cake.
Every second the clock ticks, cut the cake in half.
Make the clock variable, it ticks n times a second.
The limit clock as n tends to infinity applies an infinity of divisions to the cake in 1 second. There is no final operation.

There's nothing logically inconsistent in this, it's just not "physical". — fdrake

The lamp starts off. Every time the clock ticks a lamp turns from off to on or from on to off as applicable. Thomson's lamp shows that this leads to a logical inconsistency.

I don't see that. At best he showed that one example is undefined. To prove something impossible it must be shown that there is not a single valid one. To prove them physically possible, one must show only a single case (the proverbial black swan). Nobody has done either of those (not even Zeno), so we are allowed our opinions. — noAxioms

Take my explanation of Thomson's lamp above:

A. At t₀ the lamp is off, at t_1/2 I press the button, at t_3/4 I press the button, at t_7/8 I press the button, and so on ad infinitum

Compare with:

B. At t₀ the lamp is off, at t_1/2 I press the button

The status of the lamp at t₁ must be a logical consequence of the status of the lamp at t₀ and the button-pressing procedure that occurs between t₀ and t₁ because nothing else controls the behaviour of the lamp.

If no consistent conclusion can be deduced about the lamp at t₁ then there’s something wrong with your button-pressing procedure.

So the fact that the status of the lamp at t₁ is "undefined" given A is the very proof that the supertask described in A is impossible. — Michael

The important part is in bold. If there is a problem with the button-pressing procedure, which there is in the case of A, then this problem remains even if the button is broken and doesn't actually turn the lamp on – it turning the lamp on and off isn't the reason that the supertask is impossible but simply demonstrates that the supertask is impossible.

And this problem remains even if rather than press a broken button we recite the natural numbers or even recite a single digit on repeat.

The reason that the supertask in Thomson's lamp is impossible isn't because of what operation is performed but because of how the operation is performed: halving the time between each subsequent operation ad infinitum. This is proven impossible, and as such Thomson's lamp proves that all such supertasks are impossible.

and all you have done is taking a claim that I am obviously going to deny — Bob Ross

So you deny that “I believe that aliens exist” is a proposition?